Hi,
Have figured out your balancing problem, the solution to your question is simple math. Here is a simple explanation. The CG of a model plane is calculated using a percentage of the cord lengths, located at the root…large end of the wing and wing tip. On most planes it's about 25 to 30%, depending on the style of plane. The formula is A x B = C were A = root, B = % and C = the length from the leading edge.
Here is an example…lets say you measure the root and its 12 inches which is A.
The % in this case let’s say it’s 30% or .30…B
Multiply 12 x .30 = 3.6…(3.6 is about 3 5/8ths inch)…C
Now do the same calculation using the cord length at the wind tip.
Let’s say you measure it at 5 inch.
5 x .30 = 1.5 inch
Measure the 3 5/8 from the leading edge at the root and mark it. Measure the 1.5 from the leading edge at the wing tip and mark it. Now draw a line between the 2 points. Any were on that line is the CG of the plane.
But wait we already have the measurement (C) for the CG but it’s at the wing tip. We need the line between the 2 points.
So what do we need to solve the problem, we need B the % used to calculate lengths from the leading edge.
I’ll make it simple for you. A x B = C. To solve for B we use a little algebra magic.
The formula now C ÷ A = B. C is the given measurement for the Seagull Extra 260 is 7 to 8 cm at the wing tip. (get out your metric scale or conversion chart)
So measure the cord length at the wing tip, that is A. So now we have the A and the C… use the formula and find the % or B.
Measure the cord length at the root and multiply that length times the %. That will give you the C value or length from the leading edge at the root. Draw a line between the 2 point as I explained before. Any were on that line is the CG. Simple huh!
If you still don’t get it there is a book... Acrobatics for Everyone by Dave Patrick, he explains it. In fact if your going to be flying the Extra 260 it would be a good book to read.
Good Luck and Happy Flying
Aerosman[8D]